codeforces 835c Star sky

C. Star sky

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

The Cartesian coordinate system is set in the sky. There you can see n stars, the i-th has coordinates (xi, yi), a maximum brightness c, equal for all stars, and an initial brightness si (0 ≤ si ≤ c).

Over time the stars twinkle. At moment 0 the i-th star has brightness si. Let at moment t some star has brightness x. Then at moment (t + 1) this star will have brightness x + 1, if x + 1 ≤ c, and 0, otherwise.

You want to look at the sky q times. In the i-th time you will look at the moment ti and you will see a rectangle with sides parallel to the coordinate axes, the lower left corner has coordinates (x1i, y1i) and the upper right — (x2i, y2i). For each view, you want to know the total brightness of the stars lying in the viewed rectangle.

A star lies in a rectangle if it lies on its border or lies strictly inside it.

Input

The first line contains three integers n, q, c (1 ≤ n, q ≤ 105, 1 ≤ c ≤ 10) — the number of the stars, the number of the views and the maximum brightness of the stars.

The next n lines contain the stars description. The i-th from these lines contains three integers xi, yi, si (1 ≤ xi, yi ≤ 100, 0 ≤ si ≤ c ≤ 10) — the coordinates of i-th star and its initial brightness.

The next q lines contain the views description. The i-th from these lines contains five integers ti, x1i, y1i, x2i, y2i (0 ≤ ti ≤ 109, 1 ≤ x1i < x2i ≤ 100, 1 ≤ y1i < y2i ≤ 100) — the moment of the i-th view and the coordinates of the viewed rectangle.

Output

For each view print the total brightness of the viewed stars.

Examples

input

2 3 31 1 13 2 02 1 1 2 20 2 1 4 55 1 1 5 5

output

303

input

3 4 51 1 22 3 03 3 10 1 1 100 1001 2 2 4 42 2 1 4 71 50 50 51 51

output

3350

Note

Let's consider the first example.

At the first view, you can see only the first star. At moment 2 its brightness is 3, so the answer is 3.

At the second view, you can see only the second star. At moment 0 its brightness is 0, so the answer is 0.

At the third view, you can see both stars. At moment 5 brightness of the first is 2, and brightness of the second is 1, so the answer is 3.

这道题学会了前缀和的知识，就是给2个点分别是左上和右下，那么这个矩形里面的内容可以由一个公式得到，就是ans=dp[x2][y2]+dp[x1-1][y1-1]-dp[x1-1][y2]-dp[x2][y1-1];

但是在这之前需要去预处理dp数组，则是由dp[i][j]=dp[i][j]+dp[i-1][j]+dp[i][j-1]-dp[i-1][j-1]来得到。

#include<bits/stdc++.h>using namespace std;int ma[111][111][11];int main(){int num,que,light;while(cin>>num>>que>>light){int x,y,c;int i,j,k;memset(ma,0,sizeof(ma));for(i=1;i<=num;i++){cin>>x>>y>>c;ma[x][y][c]++;}for(i=1;i<=100;i++){for(j=1;j<=100;j++){ for(k=0;k<=light;k++)     ma[i][j][k]=ma[i][j][k]+(ma[i-1][j][k]+ma[i][j-1][k]-ma[i-1][j-1][k]);    }   }while(que--){int x1,x2,y1,y2,t;cin>>t>>x1>>y1>>x2>>y2;long long int ans=0;for(i=0;i<=light;i++){int l1=(i+t)%(light+1);ans=ans+(ma[x2][y2][i]+ma[x1-1][y1-1][i]-ma[x2][y1-1][i]-ma[x1-1][y2][i])*l1;}cout<<ans<<endl;}}return 0;}